Extracting the Square Root of a Voltage

I’m trying to discover a circuit that will produce a voltage which is some factor of the square root of the input voltage. I.e. $V_{out}(t) = Ksqrt{V_{in}(t)}$. The factor K is irrelevant.

I looked at the circuit at the bottom of this page. The problem is it uses a MOSFET and the formula predicting the output requires various parameters $mu_n, C_{ox}, V_{th}$ (some of which I imagine vary greatly even among devices of the same model, and some of which I wouldn’t know how to find from the data sheets)

I’d like to find an alternative circuit which has a consistent and predictable output, before I purchase the necessary components.

When I say K is irrelevant, I just meant that I can later amplify the output by a constant factor if needed. It needs to be consistent and predictable though.

5 Responses to “Extracting the Square Root of a Voltage”

  1. Szymon Bęczkowski says:

    An easy approach would be to use an analog multiplier (MC1495 was an early one, Analog Devices AD633 or Burr-Brown(oops, Texas Instruments!) MPY534 are better newer ones) as a squaring circuit, in the feedback loop of an op-amp.

    To use a multiplier to square a voltage simply connect that voltage to both inputs. Connect your input voltage to opamp’s non-inverting input, the opamp output to the multiply inputs and the multiply output to opamp’s inverting input.

    If $ V_{out}^2 = V_{in} $ then $ V_{out} = sqrt{V_{in}} $.


    simulate this circuit – Schematic created using CircuitLab

    Details such as DC biasing left as an exercise…

    (Side note : the analog multipliers rely heavily on “matched pairs” of transistors; it is relatively easy to match 2 transistors if you make both at once in the same area on the same chip!)

  2. dext0rb says:

    From TI application note 31:

    enter image description here

    May work with other op-amps. See the application note for details on running the LM101A off a single-ended supply.

  3. compumike says:

    if you have some BJTs plus an op-amp lying around, a quick translinear BJT analog square root is all yours! V(OUT) = SQRT(V(IN))/10 in this case:

    (Open & run the DC Sweep simulation in CircuitLab.)

    As far as “matched transistors”, in this case:

    • mismatch in Q1/Q2/Q3/Q4 or in Q6/Q7 will produce a slight scale factor error (which you’ve stated you don’t care about much anyway)
    • Q5 is not match dependent
    • variation in temperature between different transistors may produce scale error
    • you can simulate mismatch by adjusting I_S of one transistor. See this LED example for something similar in the LED case. (You can also have B_F “beta” mismatch, but in this particular circuit, it’s less of a factor.)

    I added some notes on the schematic. I’m sure others can help simplify or make this more robust, but I hope it’s a good start using parts you probably already have on your bench!

  4. user27350
    < says:
September 10, 2013 at 11:00 am

This is not intended to be an answer or an explicitic solution, rather an expanation why there aren’t any single chip integrated solutions. Perhaps the demand is too low, when you can use a digital solution now with quantization using 12 or 16 ADC’s with log codecs or log algorithms and divide by 2 in binary as the log of the exponent ^(0.5) has a 0.5 multiplier in the result.

Square root designs come in many analog variations from 1 to 16 integrated parts with complexities of precision matching, current mirrors, bias mirrors to use the quadratic non-linear behavior of FETs. They have been a perpetual research topic of EE Profs with getting controlled results spanning from 3 to 7+ decades. Problems result from variations in RgsON , Vgs threshold and self heating.

Few of these research topic experiments have ever made their way into production, perhaps because of the difficulty of controlling the process of doping and fabrication contros to get the required consistency, which are orders of magnitude more difficult than CMOS logic. The zero reference is the most critical for errors and a differential output offers more linearity in the Sq Rt result. Considering negative feedback is inverting is used, it is academic that sq. rt amps tend to take a negative input to give a positive ouput, yet this is not an maginary number. Ha.

Have fun.


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